Q. Acceleration due to gravity $g$ for a body of mass $m$ on earth’s surface is proportional to (Radius of earth = R, mass of earth = M)

A) $M/R\hat{}2\ $
B) $m^{0}$
C) $mM$
D) $1/R^{3/2}$
Explanation

(a) According of law of gravitation, the force of attraction acting on the body due to earth is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F = G\frac{Mm}{R^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
The acceleration due to gravity $g$ in the body arises due to the force $F$ from Newton’s second law of motion, we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F = mg\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ mg = G\frac{Mm}{R^{2}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{GM}{R^{2}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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